= Solution
Equal <energies> mean $dE=0$ in the linear approximation. Evaluate the coefficients at particle $A$, where $v_A=0.90c$, and use $\Delta v=0.01c$. The fractional <rest mass> change is
$$
\frac{m_B-m_A}{m_A}\approx-\frac{v_A\Delta v}{c^2-v_A^2}
=-\frac{0.90\times0.01}{1-0.90^2}=-\frac9{190}.
$$
Consequently
$$
\boxed{m_A-m_B\approx\frac9{190}m_A\approx0.0474m_A,\qquad A\text{ has the larger rest mass}.}
$$
As a check beyond the requested first-order approximation, exact equality of <relativistic energy> gives $m_B/m_A=\sqrt{(1-0.91^2)/(1-0.90^2)}=\sqrt{1719/1900}\approx0.95118$, consistent with the sign and approximate size of the difference.
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